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Message from discussion y(f(x))=y(x)+x
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Marc Nardmann  
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 More options Jan 9 2005, 8:56 am
Newsgroups: sci.math.research
From: Marc Nardmann <Marc.Nardm...@bigfoot.de>
Date: Sun, 09 Jan 2005 14:56:27 +0100
Local: Sun, Jan 9 2005 8:56 am
Subject: Re: y(f(x))=y(x)+x
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Maxim wrote:
>Is it possible to solve equation y(f(x))=y(x)+x if f(.) is a known
>function? (e.g., f(x)=exp(x)-1  ?)

If f(0)=0 and f is differentiable at 0 with f'(0)=1, as in the case
f(x)=exp(x)-1, then there is no solution y which is differentiable at 0.

PROOF: Assume that y is differentiable at 0. Then so is y°f. For x not
equal to 0, we have [(y°f)(x)-(y°f)(0)]/x - [y(x)-y(0)]/x = 1; so for x
-> 0, we obtain 1+y'(0) = (y°f)'(0) = y'(f(0)) f'(0) = y'(0), a
contradiction.

-- Marc Nardmann


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